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follow-up capacitor Q's ? Answered

I am amazed at all the great responses I got with my last question, thank you all! :D Though now I have a few more:

7) Should I try to learn how to work with complex impedance (capacitors, resistors, and inductors in all sorts of weird configurations) Also, can I treat reactance in general as a resistance when looking at capacitors in series or parallel with resistive loads and stuff? For example, can I simply add up Xc, Xi, and R for a total impedance? Or do I have to worry about phase shifts and stuff? 

8) Is it OK to say that "Q" or charge is a more theoretical physics concept and is not too important with practical electronics? (C, V, and I being more of the focus and "ignoring" Q is OK?)

9) I have added a few of the "slides" and sneak-peaks to my upcoming video. If anything is wrong don't hesitate to nitpick and point it out!

10) capacitor fall under 2 major categories, polarized and nonpolarized.

10a) [under the 'polarized' branch] Electrolytic and tantalum capacitors are used for bulk filtering, but are evil and do not tolerate overloads particularly well. Especially tantalums. They tend to be available in huge capacitances, but can be "leaky" and have high ESR and series inductance.

10b) [under nonpolarized branch] Ceramic capacitors are the most common type of capacitors, and come in a few types. Generally used for local decoupling. They are pretty robust and tolerate overloads. Film capacitors find more use in high voltage applications, have lower leakage, better high frequency performance, and certain types have self-healing properties allowing them to tolerate overloads and surges the best. Mica capacitors are generally the most stable, with the lowest leakages, so they find uses for more critical analog applications.


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Jack A Lopez
Jack A Lopez

Best Answer 5 years ago

This answer only addresses parts (7), (8), and (9), in the order (8) then (7), then (9).


Starting with question (8), the reason I mentioned Q (in my previous answer) is because it is part of the back story of how capacitors work. Q, charge, is the actual physical stuff being stored in an energized capacitor.

Q = C*vC

If you want to, you can make the variable Q go away. Recall current is just the time derivative of charge. Take the derivative of both sides, and get:

iC = dQ/dt = C*(dvC/dt)

where iC is current flowing into the capacitor.

Or just substitute, Q = integral(i*dt), to get:

vC = vC(t) = (1/C)*Q = (1/C)*integral(i*dt)

If I were to try to explain inductors, I would mention magnetic flux, Phi, the actual physical stuff being stored in an energized inductor. The defining relationship also includes N, the number of turns.

N*Phi = L*i

If you want to, you can make Phi go away, take the time derivative of both sides, and get:

vL= vL(t) = N*(dPhi/dt) = L*(di/dt)

Note, the voltage across an inductor is an induced voltage from the changing flux in the turns, and vL(t) = N*(dPhi/dt) is just Faraday's Law.

The defining relationship for a resistor, is just Ohm's law.

vR = vR(t) = R*iR(t)

To sort of summarize, I can write an equation for the voltage on a resistor, inductor, or capacitor, {vR, vL, vC} respectively, in terms of the current {iR, iL, iC} flowing through a resistor, inductor, or capacitor, like so:

vR(t) = R*i(t)

vL(t) = L*(di/dt)

vC(t) = (1/C)*integral(i*dt)

In words:

The voltage across a resistor is directly proportional to the current flowing through it. The constant of proportionality is R.

The voltage across a inductor is directly proportional to the time derivative of the current flowing through it. The constant of proportionality is L.

The voltage across a capacitor is directly proportional to the time integral of the current flowing through it. The constant of proportionality is (1/C).

So, if you want to tell the story, but without mentioning charge Q, or magnetic flux Phi, then I guess that is where you start, with:

vR(t) = R*i(t)

vL(t) = L*(di/dt)

vC(t) = (1/C)*integral(i*dt)

and those equations are axiomatic; i.e. the answer to why are they true, is because they just are. I mean, ignoring Q and Phi.

So far, all these equations are functions of time, t, and I have intentionally been using lowercase letters for these functions.

From there it is possible to invoke some unholy voodoo called a Lapace transform,


to get some related functions of a completely different variable, s (not t). Following the usual tradition, I use uppercase letters for these:

VR(s) = R*I(s)

VL(s) = L*s*I(s) = s*L*I(s)

VC(s) = (1/C)*(1/s)*I(s) = (1/(s*C))*I(s)

where {VR, VL, VC} are complex valued functions of s, which itself is complex valued. s, in terms of its real part and imaginary part,

s = sigma + j*omega

(This would look much prettier with the actual greek letters.)

Anyway, the next step is to divide both sides by I(s). Then substitute s = j*omega ( i.e sigma=0, which is essentially a useful subset of all s, called the "steady state condition"), and this gives the usual formulas for impedance,for R, L, or C, as functions of omega.

ZR = VR(j*omega)/I(j*omega) = R

ZL = VL(j*omega)/I(j*omega) = j*omega*L

ZC = VC(j*omega)/I(j*omega) = (1/(j*omega*C))

and of course, omega is angular frequency, measured in radians per second. It is related to ordinary frequency f, measured in cycles per second, by a factor of 2*pi.

omega = 2*pi*f

So basically this gives you two, widely used, methods for analyzing circuits containing resistors, inductors, and capacitors. These are time domain and frequency domain.

The trouble with time domain, (every function is a function of time) is you've got all these derivatives and integrals, with respect to time. So, to analyze a circuit in time domain, you wind up needing to solve a system of differential equations.

Frequency domain, (every function is a function of omega) is usually easier to work with, because there are no differential equations. The only difficulty is all your functions are complex valued. They have real part and imaginary part, or equivalently magnitude and phase, and it helps if you have some familiarity with what this means.


Regarding question (7), it kind of seems like you're asking about resistance, reactance, impedance... aren't they all kind of the same thing? Well, kind of, but not really.

Ohm's law defined a resistor (a lump of resistance) as this thing, for which the voltage v(t) across it is proportional to the current i(t) flowing through it, with R as the constant of proportionality.

v(t) = R*i(t) <==> R = v(t)/i(t)

Moreover it is understood, {v, i, R} are real numbers, and the equality is true in an instantaneous sense; i.e. it is true at that instant t, in time. Also t is real valued.



is defined in a similar way. Like resistance, impedance is the ratio of voltage and current, but now voltage, current, and impedance {V,I,Z} are all complex numbers, and functions of omega (angular frequency).

V(omega) = Z(omega)*I(omega)

Z(omega) = V(omega)/I(omega)

Note omega is real valued, although for some purposes it is useful to define a complex valued frequency variable, such as the Laplace transform varible, s, (shown earlier) which is essentially a complex valued frequency.

I think this believing impedance is like resistance, as defined by Ohm's Law, maybe kind of requires a leap of faith. Once you have convinced yourself there is such a thing as complex amplitude, a kind of thing consisting of both magnitude and phase,


and also that it makes sense for AC voltages, currents, and impedances to be represented in this way, then I think impedance will make a lot more sense.

By the way, it just so happens for impedors (an impedor is a lump of impedance) wired in series, the impedances add together, the same way resistors in series, the resistances add together. Zser=Z1+Z2+Z3 For impedors wired in parallel, well, it is actually the same formula as for resistors, again with Rs replaced Zs; i.e Zpar=1/((1/Z1)+(1/Z2)+(1/Z3))

The only tricky part is that all the Zs are complex numbers. The Wikipedia article on Electrical impedance


explains this, but doesn't give any super easy examples, like, maybe, you are wondering what is the impedance of a resistor R and capacitor C connected in series.

Z = ZR + ZC

Z = R + (1/(j*omega*C)) = R - j*(1/(omega*C))

Or for resistor R and capacitor C connected in parallel

Z = 1/((1/ZR)+(1/ZC)

Z = 1/((1/R) + (j*omega*C))

Z = R*(1/(1 + omega^2*R^2*C2))*(1 - j*omega*R*C)

Final note for question (7): The word "reactance" (usually denoted with capital letter Xs) is a thing more narrow than impedance. The Wiki article on Electrical Impedance, linked above, sort of explains this, but I'll explain it too. Reactance is the imaginary part of impedance. Moreover, inductors and capacitors (the ideal kind) are understood to be "pure" reactance, since the formulas for ZL = j*omega*L and ZC=1/(j*omega*C) have zero real part.

Also the word "resistance" can mean the real part of complex impedance. A resistor (the ideal kind) is thought of as "pure" resistance, since the impedance of an ideal resistor is ZR = R +j*0. The real part of ZR is R, and imaginary part or ZR is zero.


Regarding slide number 5: Q = C*V is the defining equation for capacitance, for what a capacitor is. Differentiate both sides with respect to time, and get

I(t) = dQ/dt = C*(dV/dt)

So (dV/dt) = (1/C)*I(t). So (dV/dt) is proportional to I(t) flowing into the capacitor. Moreover (dV/dt) will be small if you make C large.

In slide 4, is that a puff of magic smoke? Ha!


It's a little hard to see, but if that's what it is, hey, nice one! I mean, that is an important lesson, the lesson of what-not-to-do, i.e. what happens when ratings are exceeded, polarities reversed, etc.


Answer 5 years ago

probably a dumb question, but what is the difference between Xc and Zc? X seems to refer to reactance while Z seems to refer to impedance as a whole. Are Xc and Zc equal in ideal capacitors, that is, no inductive reactance or ESR?

Jack A Lopez
Jack A Lopez

Answer 5 years ago

Zc and Xc have the same magnitude, for sure, but they differ by a factor of j.

Or maybe it is a factor of -j? In the event Xc was defined in such a way that makes Xc real and positive, then Zc = -j*Xc. There's a blurb about this in the Wiki article on reactance, here,

This is maybe like a nit-picky thing. You probably guessed the real part of a complex number is a real number. However, it turns out the imaginary part of a complex number is also a real number.

More specifically, if Z = X + j*Y is complex, then the real(Z)=X and imag(Z)=Y, and X and Y are both real.

As an example, just to demonstrate this is the way find-the-imaginary-part-game works, I have a copy of GNU Octave here, and I ask it to tell me the real and imaginary parts of some complex number.

>>> z=(2^-0.5)*(1+1j)
z = 0.70711 + 0.70711i
>>> abs(z)
ans = 1
>>> real(z)
ans = 0.70711
>>> imag(z)
ans = 0.70711
>>> z - real(z)
ans = 0.00000 + 0.70711i

Answer 5 years ago

Wow, That is one heck of an answer! I will definitely be doing more research into laplace transforms, since I have never learned how to do them, even in calculus 3 (vector calc)! I had a friend who told me that he was fascinated with how imaginary numbers play out in the real world, and he is currently studying theoretical physics at JMU. I guess I will start to have to learn about complex number planes, ughh. At least they are easier than derivatives and integrals though.


Answer 5 years ago

You don't learn how to do them, that's one of the beautiful things about them, you can find huge tables of standard forms, and you don't actually need to evaluate an integral again to solve differential equations.


5 years ago

7.) Circuit theory is so fundamental to the practice of electronic engineering, yes or no is not even a question to be worth answering. You need to learn it and understand it, if you want to work professionally in the field, instead of dabbling. Its as central a concept to EE as adding up to maths major, or learning grammar to a language major.

No, you can't add Xc, Xl and R together - there's a reason we're talking about COMPLEX impedance. Xc and Xl can add together and cancel.

8.) No, charge is fundamental to understanding semiconductor operation.


Answer 5 years ago

So then what is an example of where Q would be important to a EE? Obviously I'm just a EE want-to-be and I know how important to really understand E theory in the profession, but at least for capacitors and stuff, are there cases where you need to deal with Q directly? It's not like you can directly measure it like current or voltage.


Answer 5 years ago

Q is important in low noise design, and lots of modeling work. Careful using Q for charge, Q also refers to the quality factor of a tuned circuit


Answer 5 years ago

Do "legit" engineers use things like LTspice to design things, or do lots of the manual by hand work with things like nodal analysis?


Answer 5 years ago

Depends on the scale and depth of the analysis, I'll use manual circuit analysis, for a design doodle, and to extract equations that give me some handle on sensitivities of certain parameters to others, but more important is to know enough about the method to see when your tools are giving you garbage.